-
Notifications
You must be signed in to change notification settings - Fork 33
/
euler-0067.cpp
138 lines (124 loc) · 4.12 KB
/
euler-0067.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
// ////////////////////////////////////////////////////////
// # Title
// Maximum path sum II
//
// # URL
// https://projecteuler.net/problem=67
// http://euler.stephan-brumme.com/67/
//
// # Problem
// By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
//
// '' 3''
// '' 7 4''
// '' 2 4 6''
// ''8 5 9 3''
//
// That is, 3 + 7 + 4 + 9 = 23.
//
// Find the maximum total from top to bottom in [triangle.txt](https://projecteuler.net/project/resources/p067_triangle.txt) (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.
//
// __NOTE:__ This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether!
// If you could check one trillion (`10^12`) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)
//
// # Solved by
// Stephan Brumme
// February 2017
//
// # Algorithm
// The algorithm and code were copied from problem 18:
//
// The main idea is to build a data structure similar to the input data:
// but instead of just storing the raw input we store the biggest sum up to this point.
//
// All data is processed row-by-row
//
// Of course, the first row consists of a single number and it has no "parents", that means no rows above it.
// Therefore the "sum" is the number itself.
// This row now becomes my "parent row" called ''last''.
//
// For each element of the next rows I have to find its parents (some have one, some have two),
// figure out which parent is bigger and then add the current input to it.
// This sum is stored in ''current''.
//
// When a row is fully processed, ''current'' becomes ''last''.
// When all rows are processed, the largest element in ''last'' is the result of the algorithm.
//
// Example:
//
// '' 1''
// '' 2 3''
// ''4 5 6''
// initialize:
// ''last[0] = 1;''
//
// read second line:
// ''current[0] = 2 + last[0] = 3''
// ''current[1] = 3 + last[0] = 4''
// copy current to last (which becomes { 3, 4 })
//
// read third line:
// ''current[0] = 4 + last[0] = 7''
// ''current[1] = 5 + max(last[0], last[1]) = 9''
// ''current[2] = 6 + last[1] = 10''
// copy current to last (which becomes { 7, 9, 10 })
//
// finally:
// print max(last) = 10
#include <iostream>
#include <vector>
#include <algorithm>
int main()
{
unsigned int tests = 1;
unsigned int numRows = 100;
//#define ORIGINAL
#ifndef ORIGINAL
std::cin >> tests;
#endif
while (tests--)
{
#ifndef ORIGINAL
std::cin >> numRows;
#endif
// process input row-by-row
// each time a number is read we add it to the two numbers above it
// choose the bigger sum and store it
// if all rows are finished, find the largest number in the last row
// read first line, just one number
std::vector<unsigned int> last(1);
std::cin >> last[0];
// read the remaining lines
for (unsigned int row = 1; row < numRows; row++)
{
// prepare array for new row
unsigned int numElements = row + 1;
std::vector<unsigned int> current;
// read all numbers of current row
for (unsigned int column = 0; column < numElements; column++)
{
unsigned int x;
std::cin >> x;
// find sum of elements in row above (going a half step to the left)
unsigned int leftParent = 0;
// only if left parent is available
if (column > 0)
leftParent = last[column - 1];
// find sum of elements in row above (going a half step to the right)
unsigned int rightParent = 0;
// only if right parent is available
if (column < last.size())
rightParent = last[column];
// add larger parent to current input
unsigned int sum = x + std::max(leftParent, rightParent);
// and store this sum
current.push_back(sum);
}
// row is finished, it become the "parent" row
last = current;
}
// find largest sum in final row
std::cout << *std::max_element(last.begin(), last.end()) << std::endl;
}
return 0;
}